A large boulder is ejected vertically upward from a volcano with an initial speed of 39.1 m/s . Air resistance may be ignored.
1. At what time after being ejected is the boulder moving at a speed 19.6 m/s upward?
2. At what time is it moving at a speed 19.6 m/s downward?
3. When is the displacement of the boulder from its initial position zero?
4. When is the velocity of the boulder zero?
5. What is the magnitude of the acceleration while the boulder is moving?
6. What is the direction of the acceleration while the boulder is moving?
Looking for both answer and formula used.
1. At what time after being ejected is the boulder moving at a speed 19.6 m/s upward?
2. At what time is it moving at a speed 19.6 m/s downward?
3. When is the displacement of the boulder from its initial position zero?
4. When is the velocity of the boulder zero?
5. What is the magnitude of the acceleration while the boulder is moving?
6. What is the direction of the acceleration while the boulder is moving?
Looking for both answer and formula used.
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YEAAAHHHHH IT'S KINEMATICS TIME!!!! WAAHOOOO!!!!!
Okay, well...
1) 0 seconds, because it's the initial velocity. Seems like kind of a duh.
2) When the boulder is thrown up, after it reaches a velocity of 0 (when it stops rising at the top and begins to fall), and is on its way down and is at its beginning point, that is when its velocity is the same as its initial. So use the kinematic V(f) = V(i) + (a)(t). You're looking for the point at which the boulder reaches the apex of its parabolic movement (when it reaches the highest it will go). You then just double the time! Isn't that neat :) So your V(f) = 0 because it will stop moving at the top. Your V(i) is 39.1 m/s. You're trying to find t, and you know (a) because it's the acceleration due to gravity, -9.8 m/s^2. So solving the equation, you'd get 3.99 ~ 4 seconds. Double that, and you find that it reaches that velocity 8 seconds after it was launched.
3) Using the same answer, 8 seconds afterwards.
4) The velocity of the boulder is zero at 4 seconds after launch, as shown by part 2.
5) The magnitude of acceleration is 9.8 m/s^2, the acceleration caused by gravity.
6) The direction of the acceleration is downwards.
Okay, well...
1) 0 seconds, because it's the initial velocity. Seems like kind of a duh.
2) When the boulder is thrown up, after it reaches a velocity of 0 (when it stops rising at the top and begins to fall), and is on its way down and is at its beginning point, that is when its velocity is the same as its initial. So use the kinematic V(f) = V(i) + (a)(t). You're looking for the point at which the boulder reaches the apex of its parabolic movement (when it reaches the highest it will go). You then just double the time! Isn't that neat :) So your V(f) = 0 because it will stop moving at the top. Your V(i) is 39.1 m/s. You're trying to find t, and you know (a) because it's the acceleration due to gravity, -9.8 m/s^2. So solving the equation, you'd get 3.99 ~ 4 seconds. Double that, and you find that it reaches that velocity 8 seconds after it was launched.
3) Using the same answer, 8 seconds afterwards.
4) The velocity of the boulder is zero at 4 seconds after launch, as shown by part 2.
5) The magnitude of acceleration is 9.8 m/s^2, the acceleration caused by gravity.
6) The direction of the acceleration is downwards.